1) f(x) =x+1
x³ + x² + x + 1 = x + 1 <=> x³ + x² = 0
<=> x²(x+1)=0
<=> x²= 0 ou x+1 = 0
<=> x = 0 ou x = -1 S = {-1 ; 0}
2) f(x) = 0 on utilise la forme factorisée de f(x)
(x+1)(x²+1) = 0 <=> x+1 = 0 ou x²+1 = 0
x²+1 = 0 n'a pas de solution (x²+1≥1)
(x+1)(x²+1) = 0 <=> x=-1 S = {-1}
3) f(x) = g(x) on utilise la forme factorisée de f(x)
(x+1)(x²+1) = (x+1)³ <=> (x+1)(x²+1) - (x+1)³ = 0
<=> (x+1)[(x²+1)-(x+1)²] = 0
<=> (x+1)(x²+1 -x²-2x-1) =0
<=> (x+1)(-2x) =0
<=> x+1 = 0 ou -2x = 0
<=> x=-1 ou x=0 S = {-1;0}
